POJ 1019 Number Sequence

 Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20096 Accepted: 5271

Description

A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another.
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 ≤ i ≤ 2147483647)

Output

There should be one output line per test case containing the digit located in the position i.

Sample Input

283

Sample Output

22

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

题目大意：给出串“11212312341234512345612345671234567812345678912345678910123456789101112345678910……”问第N个是几

题目分析：如下

               1

               1 2

               1 2 3

               1 2 3 4

               ……

即可求出第N个在第几行 于是转化为 求串“12345678910111213……”中第N'是几

代码如下：

#include <iostream>#include <cmath>using namespace std;long long a[31270],b[31270];int main(){long long n,i,j;a[1]=b[1]=1;for(i=2;i<31270;i++){if(i<10)a[i]=a[i-1]+1;else if(i<100)a[i]=a[i-1]+2;else if(i<1000)a[i]=a[i-1]+3;else if(i<10000)a[i]=a[i-1]+4;else a[i]=a[i-1]+5;b[i]=b[i-1]+a[i];}cin>>n;while(n--){cin>>j;i=0;while(b[i]<j)i++;j-=b[i-1];i=0;while(a[i]<j)i++;i=i/int(pow(10.0,int(a[i]-j)))%10;cout<<i<<endl;}return 0;}